Exams › NEET › Physics

10 gm of ice cubes at 0°C are released in a tumbler (water equivalent 55 g) at 40°C. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L = 80 cal/g)

1. 31°C
2. 22°C
3. 19°C
4. 15°C

Correct answer: 22°C

Solution

The heat required to melt the ice is Q1 = mL = 10 × 80 = 800 cal. The heat released by the water and tumbler as they cool down to 0°C is Q2 = (mass of water + water equivalent of tumbler) × specific heat × temperature drop = (10 + 55) × 1 × (40 - T). Equating Q1 and Q2, solving gives T ≈ 22°C.

Related NEET Physics questions

• An instrument used to measure high temperature is
• Ocean currents are caused by the setting up of in water
• For the same mass, which one of the following has the maximum thermal capacity?
• The excess temperature of a body falls from \( 12^{\circ} \mathrm{C} \) to \( 6^{\circ} \mathrm{C} \) in 5 minutes, then the time to fall the excess temperature from \( 6^{\circ} \mathrm{C} \) to \( 3^{\circ} \mathrm{C} \) is (assume the Newton's cooling is valid)
• Bunty mixed 440 gm of ice at \( 0^{\circ} \mathrm{C} \) with \( 540 \mathrm{gm} \) of water at \( 80^{\circ} \mathrm{C} \) in a bowl. Then what would remain after sometime in the bowl?
• Mercury boils at \( 356^{\circ} \mathrm{C} \). However mercury thermometers are made such that they can measure temperatures up to \( 500^{\circ} \mathrm{C} \). This is done by :

⚔️ Practice NEET Physics free + battle 1v1 →